MATH2020 Algebraic Structures 1, Academic Year 2012-13,Semester 1

Notes of a lecture course by P M Schuster

Contact Details

Lecturer: Dr Peter M SchusterE-mail: pschust@maths.leeds.ac.ukOffice: School of Mathematics, 8.17eModule Webpage: http://www1.maths.leeds.ac.uk/∼pschust/teaching/algstr1/algstr1.html

Reading List

Groups

R. B. J. T. Allenby, Rings, fields, and groups, 2nd ed, Edward Arnold, 1991.M. A. Armstrong, Groups and symmetry, Springer-Verlag, 1988.G. Birkhoff and S. MacLane, A survey of modern algebra, Macmillan.Peter J. Cameron, Introduction to Algebra, Oxford University Press, 1998.B. L. Johnston and F. Richman, Numbers and Symmetry, CRC Press 1997.C. R. Jordan and D. A. Jordan, Groups, Edward Arnold, 1994.W. R. Ledermann and A. J. Weir, Introduction to group theory, 2nd edition, Longman, 1996.

Vector Spaces

M. Anthony and M. Harvey, Linear Algebra: Concepts and Methods, Cambridge University Press,2012.R. Kaye, R. Wilson, Linear algebra, Oxford University Press, Oxford, 1998.S. Lipschutz, Linear Algebra (actual title is Schaum’s outline of theory and problems of linearalgebra), McGraw-Hill, 1991.S. I. Grossman, Elementary linear algebra, Saunders College Publishing, 1994.

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A.O. Morris, Linear algebra : an introduction, Van Nostrand Reinhold (U.K), 1982.R.B.J.T. Allenby, Linear algebra, Edward Arnold, 1995.

Further reading

Bourbaki, Nicolas: Algebra. I. Chapters 1-3. Translated from the French. Reprint of the 1989English translation. Springer-Verlag, 1998.Curtis, Charles W.: Linear algebra. An introductory approach. Corrected reprint of the 1984fourth ed. Springer-Verlag, New York, 1993.Hoffman, Kenneth; Kunze, Ray: Linear algebra. Second edition. Prentice-Hall, Inc., EnglewoodCliffs, NJ, 1971.Janich, K.: Linear algebra. Springer-Verlag, New York, 1994.Derbyshire, John: Unknown quantity. A real and imaginary history of algebra. Joseph HenryPress, Washington, DC, 2006.

Coursework- worth 15% of the final mark

Coursework cannot be handed in at lectures or workshops. It has to be put instead, before thebeginning of the lecture on the day on which the coursework is due, into the appropriate marker’sgrey locker next to the UG office on level 8 of the School of Mathematics. To find out which markeryou have been assigned to please look online, or in the red handbook on display near the lockers.Missed or late coursework will be handled according to the Policy on missed or late coursework.Marked coursework will first be returned at lectures and then put into the appropriate blue trayopposite of the grey lockers, where also spare copies of handouts such as problem sets and modelsolutions will be put.

Introduction

Algebraic structures provide one way of encapsulating mathematical properties so that they canbe more easily studied, and often lead to unifying diverse areas of mathematics. In this modulewe will study two such structures, those of groups and vector spaces.

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Part I

Groups

1 Basic Definitions

Group theory may be regarded as an abstract study of symmetry, and plays a central role inmathematics and its applications. For example, the degree of symmetry of a geometrical figuremay be captured by the corresponding group, which tells us not just how many symmetries thereare, but also precisely how they interact (the ’structure’ of the group).The concept of a group comes from two separate sources:

• Symmetries, in areas of physics, and

• Solving equations of fifth degree or higher, as such equations are unsolvable systematically.

Definition: A group is a triple (G, ◦, e) where:

i) G is a set,

ii) e is an element of G,

iii) ◦ : G×G→ G, (x, y) 7→ x ◦ y, is a map.

such that, writing xy for x ◦ y:

1) x(yz) = (xy)z for all x, y, z ∈ G,

2) xe = x = ex for all x ∈ G,

3) For every x ∈ G there exists y ∈ G with xy = e = yx.

Now let (G, ◦, e) be a group. We observe the following facts:

a) For all x, a, b ∈ G, if xa = xb or ax = bx, then a = b, (cancellation property),

b) By 2), e is uniquely determined; e is the neutral element of G,

c) The y ∈ G as in 3) is uniquely determined by x ∈ G, and y is called the inverse, x−1, of x,

d) For all x, y ∈ G, (xy)−1 = y−1x−1 and (x−1)−1 = x.

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e) For x ∈ G and n ∈ N one can define xn ∈ G by x0 := e, and xn+1 = (xn)x. Thenxn+m := xnxm and xnm = (xn)m for all n,m ∈ Z where x−n = (xn)−1 for n ≥ 0.

Proof. We will prove the first three facts, and leave the rest as exercises:

a) If, say xa = xb and y is as in 3), then

a = ea = (yx)a = y(xa) = y(xb) = (yx)b = eb = b.

Hence a = b.

b) Suppose that e′ ∈ G such that xe′ = x = e′x for all x. Then e = e′e, and e′ = e′e by 2).Then e′ = e.Hence e is unique.

c) , d) and e) have been left as exercises.Hint: c) is similar to a) and b), and d) follows from c). For e) use proof by induction, byfixing one of m,n and use induction on the other.

Definition: A group G is (commutative or) abelian if xy = yx for all xy. Then one oftenwrites x+ y for xy, 0 for e, and −x forx−1.Convention: If G is abelian then write n · x for xn, where x ∈ G and n ∈ Z.

Note: ◦ can mean multiplication (·) or addition (+) or something else.

• For multiplication use e = 1 and x−1 = 1x,

• For addition use e = 0 and x−1 = −x,

• Associativity x(yz) = (xy)z holds in any group whatsoever,

• Commutativity xy = yx only holds in abelian groups e.g. GMn(K) is not abelian if n > 1.

Examples:

0) (N,+, 0) is not a group as it does not satisfy condition 3).

1) (Z,+, 0) is a group. Furthermore it is an abelian group.

2) For K ∈ {Q,R,C} set K∗ = K\{0}. Then (K∗, ·, 1) is an abelian group.

3) Given any set X, let Sym(X) be the set of all the bijections σ : X → X. This is a group withcomposition of maps σ ◦ τ [where σ ◦ τ(x) = σ(τ(x))], identity idx [where idx(x) = x] andinverse σ−1. This is called the symmetric group. Sym(X) is abelian if and only if |x| ≤ 2.

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4) Let GLn(K) (the general linear group) be the set of invertible (n × n)-matrices withentries from K. Then GLn(K) is a group with matrix multiplication, identity matrix andinverse matrix.Also GLn(K) is abelian if and only if n = 1.

2 Homomorphisms

Definition Let (G, ◦, e) and (G′, ◦′, e′) be groups. A map f : G → G′ is a homomorphism iff(x ◦ y) = f(x)◦′f(y) for all x, y ∈ G. If in addition f is bijective, then f is an isomorphism.

Lemma: Let f : G→ G′ be a homomorphism. Then:

a) f(e) = e′, and f(x)−1 = f(x−1) for every x ∈ G,

b) If f is bijective then f−1 is a homomorphism.

Proof. a) Nowe′f(e) = f(e) = f(ee) = f(e)f(e)

so e′ = f(e) by cancellation.Furthermore f(x)f(x−1) = f(xx−1) = f(e) = e′ from above. Likewise f(x−1)f(x) = e′.Hence f(x−1) = f(x)−1.

b) Set f ′ = f−1, so we have f ′ : G′ → G.

Let x′, y′ ∈ G′ and set x = f ′(x′), y = f ′(y′).Then x, y ∈ G and f(x) = x′, f(y) = y′. Now:

f ′(x′y′) = f ′(f(x)f(y)) = f ′(f(xy)) = xy = f ′(x′)f ′(y′).

Hence f ′ is a homomorphism.

Examples:

0) For K ∈ {Q,R,C}, see the determinant as a map:

det : GLn(K)→ K∗, A 7→ det(A).

This is a homomorphism because det(AB) = det(A)det(B). The determinant is also surjec-tive (as if λ = K then λ = det(A) where A is a matrix with λ in the first entry, 1 in all otherdiagonal entries and 0 elsewhere). Furthermore, the determinant is an isomorphism preciselywhen n = 1 (If n = 1 then det−1(λ) = (λ), a 1× 1 matrix, for every λ ∈ K∗).

1) For K ∈ {Q,R,C}, |det|, det|det| : GLn(K)→ K∗ are homomorphisms.

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2) For any groups G,G′

the constant map f : G→ G′, x 7→ e

′is the trivial homomorphism.

3) g : (R,+, 0)→ (R∗, ·, 1), x 7→ ex andh : (R,+, 0)→ (C∗, ·, 1), t→ e2πit = cos 2πt+ i sin 2πtare homomorphisms, because ex+y = exey, and e2πi(t+s) = e2πite2πis for any x, y, t, s ∈ R.Furthermore, g is injective, but not surjective, and h is neither injective nor surjective.

3 Subgroups

Definition: Let (G, ◦, e) be a group. A subset H of G is a subgroup if

i) e ∈ H,

ii) If x ∈ H and y ∈ H, then x ◦ y ∈ H,

iii) If x ∈ H, then x−1 ∈ H.

Then (H, ◦, e) is a group and the inclusion i : H → G, x 7→ x is an injective homomorphism.Also for every homomorphism f : G→ G

′:

1) The kernel Ker(f) = {x ∈ G : f(x) = e′} is a subgroup of G.

2) The image Im(f) = {f(x) : x ∈ G} is a subgroup of G.

Proof. 1) Kernel:

i) f(e) = e′ ⇒ e ∈ Ker(f),

ii) If x, y ∈ Ker(f) then f(xy) = f(x)f(y) = e′e′= e

′. So xy ∈ Ker(f),

iii) If x ∈ Ker(f) then f(x−1) = f(x)−1 = (e′)−1 = e

′. so x−1 ∈ Ker(f).

2) Image:

i) e′= f(e)⇒ e

′ ∈ Imf ,

ii) If x′, y′ ∈ Im(g), i.e. x

′= f(x), y

′= f(y) for some x, y ∈ G, then x

′y′

= f(x)f(y) =f(xy) ∈ Im(f),

iii) If x′ ∈ Im(f) i.e. x

′= f(x) for some x ∈ G, then (x

′)−1 = f(x)−1 = f(x−1) ∈ Im(f).

Examples:

0) {e} and G are the trivial subgroups of G,

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1) The centre Z(G) of G, (i.e. Z(G) = {x ∈ G : ∀y ∈ G(xy = yx)}) is a subgroup of G and Gis abelian if and only if Z(G) = G (left as an exercise).

2) With g : (R,+, 0) → (R∗, ·, 1), x 7→ ex and h : (R,+, 0) → (C∗, ·, 1), t → e2πit as above,(0,+∞) =Im(g) is a subgroup of (R∗, ·, 1) and {z ∈ C : |z| = 1} =Im(h) is a subgroup of(C∗, ·, 1). Furthermore, Ker(g) = {0}, and Ker(h) = {t ∈ R : e2πit = 1} = Z, which clearlyare subgroups of (R,+, 0).

3) Let K ∈ {Q,R,C}, n ≤ 1. Then SLn(K) = {A ∈ GLn(K) : det(A) = 1} (the special lineargroup of K) is a subgroup of GLn(K), for SLn(K) is the kernel of det : GLn(K) → K∗,which is a homomorphism.

Lemma: Let (G, ◦, e) and (G′, ◦′ , e′) be groups, and f : G→ G

′be a homomorphism. Then:

a) f is surjective if and only if Im(f) = G′,

b) f is injective if and only if Ker(f) = {e}.

Proof. a)

f is surjective ⇔ for every x′ ∈ G′ there exists x ∈ G such that x

′= f(x)

⇔ Im(f) = G′.

b) Recall f(e) = e′, and that f is a homomorphism if f(xy) = f(x)f(y) for all x, y ∈ G.

i) We assume that f is injective. It is enough to show that if f(x) = e′

then x = e as ifx ∈Ker(f) then f(x) = e

′. If f(x) = e

′then f(x) = f(e) and so x = e as f is injective.

Hence Ker(f) = {e}ii) We assume that Ker(f) = {e}. If f(x) = f(y) with x, y ∈ G, then

f(y−1x) = f(y−1)f(x) = f(y)−1f(x) = f(y)−1f(y)

by hypothesis. Hence f(y−1x) = e′, and so y−1x = e, i.e. x = y, hence f is injective.

Lemma:

a) For every group G, the identity map idG : G→ G, x 7→ x is a homomorphism.

b) If α : G→ G′, β : G

′ → G′′

are homomorphisms then also β ◦ α : G→ G′′, x→ β(α(x)) is a

homomorphism.

Proof. a) Is obvious.

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b) We have:β ◦ α(xy) = β(α(xy)) = β(α(x)α(y)), as α is a homomorphism,= β(α(x))β(α(y)), as β is a homomorphism,= β ◦ α(x)β ◦ α(y)Hence β ◦ α is a homomorphism.

4 Cayley Tables

Definition: The group table or Cayley table of G is a square matrix that describes the groupoperation as follows:

◦ e x ye e x yx x x ◦ x x ◦ yy y y ◦ x y ◦ y

Example: G = {0, 1, 2, 3} such that x ◦ y = (x+ y)mod4, and e = 0:

0 1 2 30 0 1 2 31 1 2 3 02 2 3 0 13 3 0 1 2

In fact G is an abelian group. Furthermore the group table of a group G is symmetric if and onlyif G is abelian. We can also use this table to find the inverses of elements.

Theorem (The Latin Square Property): In the Cayley table of a group G every element z ∈ Goccurs exactly once in each row and in each column.

Proof. Let `x : G → G, u 7→ xu, (left multiplication) and ry : G → G, v 7→ vy, (right multiplica-tion) for any x, y ∈ G. Now `x, ry are bijections with inverse `−1

x = `x−1 , r−1y = ry−1 (for example,

`x`x−1 = `x(x−1u) = x(x−1u) = (xx−1)u = eu = u, so `x ◦ `x−1 = idG). Neither `x nor ry is an

isomorphism unless x = e of y = e respectively. Now each row is the range of lx for some x ∈ Gand each column is the range of ry for some y ∈ G.

Hence, if for exampleG = {a1, a2, ..., an} then a row is aia1, aia2, ..., aian and a column is a1aj, a2aj, ..., anaj.Also to find the inverse x−1 of any x ∈ G, look at the row (or column) starting with x and picky ∈ G above (or to the left of) the only e that occurs in this row (or column).

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5 Generators

Let U,H be subgroups of a group G. Clearly U ∩ H is a subgroup of G (actually U ∩ H is asubgroup of U and H). In general, U ∪H is not a subgroup of G.

Example: Let E = {−1, 1} and P = (0,∞). then E and P are subgroups of (R∗, ·, 1) but theirunion E ∪ P = {−1, 1} ∪ (0,∞) is not a subgroup of R∗. For instance −1 ∈ E, 2 ∈ P but−1 · 2 = −2 6∈ E ∪ P .

Definition Let S be a non-empty subset of a groupG. The set 〈S〉 of all the possible x±11 x±1

2 x±13 ...x±1

m

with m ∈ Z+ and xk ∈ S for every k ∈ {1, 2, ...,m} is a subgroup of G. We have e ∈ 〈S〉 there isx ∈ S for which e = xx−1 ∈ 〈S〉. This 〈S〉 is the subgroup generated by S.If S = {a1, ..., am} then write 〈a1, ..., am〉 for 〈{a1, ..., am}〉. In particular 〈a〉 means 〈{a}〉 for which〈a〉 = {al : l ∈ Z}. Note that 〈a〉 = 〈a−1〉 for any a ∈ G.

If G is abelian, then 〈a1, ..., am〉 = {a`11 . . . a`mm : `1, . . . , `m ∈ Z} (prove this!). If, in addition, G isseen as (G,+, 0), then one writes ` · x or simply `x in place of x`, for any x ∈ G and ` ∈ Z. Hence〈a1, . . . , am〉 = {`1a1 + . . .+ `mam : `1, . . . , `m ∈ Z}; in particular 〈a〉 = {`a : ` ∈ Z}.

Examples:

0) 〈e〉 = {e}. In fact 〈x〉 = e⇔ x = e.

1) In (Z,+, 0), 〈2〉 = {2` : ` ∈ Z} = {z ∈ Z : z is even} is the set of all even integers.

2) In (Z,+, 0) we have 〈2, 3〉 = Z. In fact, H = 〈2, 3〉 is a subgroup of Z with 2, 3 ∈ H. Moreprecisely 〈2, 3〉 = {2k + 3l : k, l ∈ Z}; so, for example, 1 = 3 − 2 = 2(−1) + 3(1) ∈ H. Andthus all integers are in H as l = l · 1 = 1 + 1 + ...+ 1.

6 Order

Let G be a group, and x ∈ G. The order of G is |G| i.e. the number of elements in G if G is finite,or an unspecified infinite if G is infinite. The order of x is ord(x) = |〈x〉|. Since 〈x〉 ⊆ G, we haveord(x) ≤ |G| if G is finite, and so ord(x) is finite in this case.

Examples:

0) If G is any group then ord(x) = 1 if and only if x = e (proof left as an exercise).

1) In (Z,+, 0), every non-zero element m has infinite order, because 〈m〉 = {` ·m : ` ∈ Z} ={...,−3m,−2m,−m, 0,m, 2m, 3m, ...} is an infinite set.

Corollary: If |G| = n and x ∈ G then G = 〈x〉 precisely when ord(x) = n.

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Proof. Left as a very simple exercise.

Examples:

1) For (Q∗, ·, 1) and (R∗, ·, 1), ord(1) = 1 and ord(−1) = 2, but all other elements have infiniteorder.

2) In C∗ the elements of finite order are the roots of unity (i.e. the z ∈ C for which there ism ≥ 1 such that zm = 1). For example, ord(1) = 1, ord(−1) = 2, ord(±i) = 4, ord(w) = 3,

where w = e2πi3 = cos 2π

3+ i sin 2π

3.

Theorem Let G be any group, and x ∈ G. The following are equivalent:

i) ord(x) is finite,

ii) There are k, ` ∈ Z with xk = x` but k 6= `,

iii) There is m ≥ 1 such that xm = e.

In this case, 〈x〉 = {e, x, x2, ..., xord(x)−1}, and ord(x) = min{m ≥ 1 : xm = e} (i.e. xord(x) = e, and

if xm = e for any m ≥ 1 then ord(x) ≤ m). Also, for every m ≥ 1, xm = e(†)⇔ m0|m.

Proof. i) ⇒ ii) If 〈x〉 = {x` : ` ∈ Z} is finite, then there must be k 6= ` with xk = x` becauseotherwise 〈x〉 would be infinite.

ii) ⇒ iii) If xk = x` with k < `, then m = `− k ≥ 1 satisfies xm = e as required.

iii) ⇒ i) Set m0 = min{m ≥ 1 : xm = e}. For 0 ≤ i < j < m0 we have 1 ≤ j − i < m0 and thusxj−i 6= e (by the definition of m0), i.e. xi 6= xj. So, the e, x, x2, ..., xm0−1 are all distinct.We next show that 〈x〉 = {e, x, x2, ..., xm0−1}. By the definition of 〈x〉, we have ⊇. As for ⊆,if y ∈ 〈x〉, i.e. y = xl for some l ∈ Z, then write l = m0q + r where 0 ≤ r < m0, for whichy = x` = (xm0)qxr = xr ∈ {e, x, x2, ..., xm0−1} as required. So 〈x〉 = {e, x, x2, ..., xm0−1}.Since the e, x, x2, ..., xm0−1 are all distinct, it follows that ord(x) = m0 is finite.

We thus have shown that i), ii), iii) are equivalent; that ord(x) = min{m ≥ 1 : xm = e}; and that〈x〉 = {e, x, x2, ..., xm0−1}. It remains to verify (†), of which ⇐ is clear from xm0 = e. As for ⇒,assume that xm = e, and write m = m0s + t with 0 ≤ t < m0. Since e = xm = (xm0)sxt = xt, wehave t 6≥ 1, i.e. t = 0, which is to say that m0 divides m.

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7 Cyclic Groups

Definition: A group G is called cyclic if there is x ∈ G with G = 〈x〉. Every cyclic group isabelian and is countable (i.e. is finite, or has a one-to-one correspondance with N).

Proof. A cyclic group has xkx` = xk+` = x`+k = x`xk, hence is abelian.

Examples:

0) None of Q∗,R∗,C∗ are cyclic, and neither is the circle {z ∈ C : |z| = 1} as a subgroup of C∗.

1) For each n ≥ 1, the nth roots of unity form the subgroup {z ∈ C : zn = 1} of (C∗, ·, 1), a

cyclic group of order n, for example generated by w = e2πin or w = e

−2πin , where w is the

complex conjugate of w, as 〈w〉 = 〈w〉.

2) (Z,+, 0) is a cyclic group of infinite order, and Z = 〈a〉 ⇔ a ∈ {−1, 1}.

3) For each n ≥ 1, let Zn = {0, 1, ..., n − 1}, and let + : Zn × Zn → Zn denote +modn, i.e.(x, y) 7→ r precisely when x+ y = qn+ r where q ∈ Z and r ∈ Zn. Then (Zn,+, 0) is a cyclicgroup of order n, in which −x is n− x, and Zn = 〈1〉 = 〈n− 1〉.

Theorem: Let G be a group, x ∈ G and n, k ≥ 1. If ord(x) = n then ord(xk) = ngcd(k,n)

where

gcd(k, n) is the greatest common divisor of k and n.

Proof. This proof is omitted, as it is similar to a previous proof, using division with remainder.

Corollary: Let G be a group.

a) If G is a cyclic group, G = 〈x〉, and |G| = n, then for every k ≥ 1, G = 〈xk〉 if and only ifk, n are coprime i.e. gcd(k, n) = 1.

b) If G is cyclic and |G| is a prime number then G = 〈y〉 for every y ∈ G\{e}.

Proof. a) We already know that if |G| = n, then G = 〈xk〉 ⇔ ord(xk) = n. From the theoremwe see that ord(xk) = n⇔ gcd(k, n) = 1. Also, gcd(k, n) = 1⇔ k, n coprime.

b) Left as an exercise. Hint: Look at G = 〈x〉.

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Examples: In Zn = {0, 1, ..., n− 1}, ord(k) = ord(k · 1) = ngcd(k,n)

:

Z3 = 〈1〉 = 〈2〉, which all have order 3.

Z4 = 〈1〉 = 〈3〉, which all have order 4, but 〈2〉 = {0, 2} 6= Z4, and 2 has order 2.

Z6 = 〈1〉 = 〈5〉, which all have order 6, but 〈2〉 = 〈4〉 = {0, 2, 4} 6= Z6, and 〈3〉 = {0, 3} 6= Z6,

where 2, 4 have order 3, and 2 has order 3.

Definition: Two groups G,G′

are isomorphic if there is an isomorphism G→ G′.

For short G ∼= G′. This is an equivalence relation (i.e. the relation is reflexive, transitive and

symmetric).

Lemma: If G,G′

are isomorphic, then |G| = |G′ |, and G is abelian (cyclic) if and only if G′

isabelian (cyclic).

Proof. Left as an exercise.

Examples:

0) Zn 6∼= Zk if n 6= k as they have different orders.

1) ({−1, 1}, ·, 1) ∼= Z2 = ({0, 1},+, 0) (workshop 1).

2) (R,+, 0) ∼= ((0,∞), ·, 1) (consider exp/log functions)

Theorem (Classification of cyclic groups): Let G be a cyclic group. Then:

a) If |G| is infinite then G ∼= Z.

b) If |G| = n then G ∼= Zn.

More precisely, if G = 〈x〉 and f(`) = x` then in case a) f : Z→ G is an isomorphism, in case b)f : Zn → G is an isomorphism.

Proof. a) Since G = 〈x〉 = {x` : ` ∈ Z}, we have G = Im(f); whence f is surjective. Alsof is injective because xk 6= x` for k 6= l since ord(x) = |〈x〉| = |G| is infinite. Furtherf(k + `) = xk+` = xkx` = f(k)f(`) and therefore f is a homomorphism. In all, G ∼= Z

b) As G = 〈x〉 = {xl : l ∈ Zn}, f is surjective. Since |Zn| = n = |G|, f is even bijective. Itremains to show that f(k ⊕ `) = xkx` where ⊕ is addition in Zn. i.e.:

k ⊕ ` =

{k + ` if k + ` < n,k + `− n if k + ` ≥ n.

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Since ord(x) = n and thus xn = e, we have

f(k ⊕ `) =

{xk+` = xkx` if k + ` < n,xk+`−n = xkx`x−n = xkx` if k + ` ≥ n,

and so f(k ⊕ `) = xkx` = f(k)f(`). Hence f is a homomorphism and so G ∼= Zn.

Examples: Let n ≥ 2.

1) Let Gn = {z ∈ C : zn = 1} be the set of all nth roots of unity. As a subgroup of (C∗, ·, 1), this

Gn is a cyclic group of order n, i.e. Gn∼= Zn, and α : Zn → Gn, k 7→ w, where w = e

±2πin , is

an isomorphism, because Gn = 〈w〉.

2) Let Hn be the set of all rotations of a regular polygon Pn in C with n vertices. View Hn as asubgroup of (Sym(Pn), ◦, idPn). Then Gn

∼= Hn via the isomorphism β : Gn → Hn, wk 7→ ρk,

where ρ is the (anti-) clockwise rotation of Pn through 2πn

.

So Zn∼= Hn via β ◦ α : Zn

α−→ Gnβ−→ Hn:

(a) Zn (b) Gn (c) Hn

Figure 1: Three isomorphic groups

8 Direct Products

Definition: If G1 and G2 are groups with neutral elements e1 and e2, then the cartesian productG1 × G2 = {(x1, x2) : x ∈ G1, x2 ∈ G2} is a group with (x1, x2)(y1, y2) = (x1y1, x2y2) ande = (e1, e2); this group is the direct product G1 ×G2 of G1 and G2.Fact: G1 ×G2 is abelian if and only if G1 and G2 are abelian.

Examples:

1) (C,+, 0) ∼= (R,+, 0)× (R,+, 0),a+ ib ! (a, b).

13

2) (C∗, ·, 1) ∼= ((0,+∞), ·, 1)× (C, ·, 1),re2πit ! (r, e2πit),where the unit circle C = {z ∈ C : |z| = 1} is, as usual, seen as a subgroup of (C∗, ·, 1).

Lemma: Let G1, G2 be finite groups.

1) |G1 ×G2| = |G1| · |G2|

2) ord((x1, x2)) = lcm(ord(x1), ord(x2)) for x1 ∈ G1, x2 ∈ G2, where lcm(a, b) is the leastcommon multiple of a and b.

Proof. (a) is clear. (b) is left as an exercise.

Theorem: Let G1, G2 be cyclic groups, where |G1| = n1, |G2| = n2. Then G1×G2 is cyclic if andonly if n1, n2 are coprime. In this case if G1 = 〈x1〉, G2 = 〈x2〉 then G1 ×G2 = 〈(x1, x2)〉.

Proof.

〈(x1, x2)〉 = G1 ×G2

⇔ ord((x1, x2)) = |G1 ×G2|⇔ lcm(ord(x1), ord(x2)) = n1 · n2

⇔ gcd(n1, n2) = 1

⇔ n1, n2 are coprime.

We have used once more that lcm(n1, n2)gcd(n1, n2) = n1n2.

Examples:

1) Z2 × Z3∼= Z6; Z5 × Z6

∼= Z30. In fact, Z2 × Z3 and Z5 × Z6 are cyclic of order 6 and 30,respectively.

2) Z2 × Z2 6∼= Z4; Z2 × Z4 6∼= Z8. In fact, Z2 × Z2 and Z2 × Z4 are not cyclic.

9 Lagrange’s Theorem

Definition: Let G be a group and H a subgroup. The coset (more precisely, the left cosetmodulo H) of x ∈ G is xH = {xz : z ∈ H}. If G is abelian, then one often writes x + H ={x+ z : z ∈ H}.

Examples:

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1) G = (R,+, 0), H = Z ⇒ x + Z = {..., x − 2, x − 1, x, x + 1, x + 2, ...}, e.g. 13

+ Z ={..., −5

3, −2

3, 1

3, 4

3, ...}. Also, notice that 0 + Z = Z.

2) G = (C∗, ·, 1), with subgroup C = {z ∈ C : |z| = 1}. Then xC = {xeit : t ∈ R} is the circlewith radius |x| and centre 0. Note that C∗ =

⋃x∈C∗ xC, and that 1C = C.

Lemma: Let G be a group, and H a subgroup of G. For x, y ∈ G the following statements areequivalent:

i) xH = yH,

ii) xH ⊆ yH,

iii) x ∈ yH,

iv) y−1x ∈ H,

v) xH ∩ yH 6= ∅.

Proof. i) ⇒ ii) ⇔ iii) ⇒ iv) are not hard to verify.iv) ⇒ v): If y−1x ∈ H, then:

x = ex = (yy−1)x = y(y−1x) ∈ yH.

Since also x = xe, and e ∈ H , we have x ∈ xH as well. In all, x ∈ xH ∩ yH.v)⇒ i): If xH ∩ yH 6= ∅, then xz1 = yz2 for some z1, z2 ∈ H. Hence x = y(z2z

−11 ) ∈ yH, and thus

xH ⊆ yH. The method to show xH ⊇ yH is similar.

Corollary:

a) x ∈ xH,

b) xH = H ⇔ x ∈ H.

Definition: Let G be a finite group and H ⊆ G a subgroup. Then the quotient set G/H ={xH : x ∈ G} is finite, and |G/H| is the index of H in G. Sometimes: [G : H] is written for|G/H|, and the quotient set of the index are sometimes specified by saying ’G modulo H’.

Theorem (Lagrange’s Theorem): If H is a subgroup of a finite group G, then

|G| = |G/H| · |H| .

The conclusion of Lagrange’s Theorem can be put suggestively as

|G/H| = |G|/|H| .

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Proof. Write G/H = {x1H, ..., xmH} where G/H = m, so if i 6= j then xiH 6= xjH, i.e. xiH ∩xjH = ∅ from the above lemma. Further, G = x1H∪ ...∪xmH, because if x ∈ G, then xH ∈ G/H,and thus xH = xkH for some k ≤ m. As we thus have a disjoint union,

|G| = |x1H|+ |x2H|+ ...+ |xmH| .

For each k ≤ m the map H → xkH, u 7→ xku is a bijection with inverse map xkH → H, v 7→ x−1k v,

so |xkH| = |H| for every k ≤ m. In all |G| = m|H| = |G/H||H| as required.

Corollary: Let G be a finite group, H ⊆ G a subgroup, and x ∈ G. Then:

a) |H| divides |G|,

b) ord(x) divides G.

c) x|G| = e.

Proof. Parts a) and b) are consequences of the Theorem; recall that ord(x) = |〈x〉| is the order ofa subgroup. For part c) recall that xord(x) = e. Then, by b), kord(x) = G for some k ≥ 1. Nowx|G| = xkord(x) = (xord(x))k = ek = e.

Example: G = Z6 = {0, 1, 2, 3, 4, 5} with addition mod 6, and 0 as the neutral element.

a) If H = 〈2〉 = {0, 2, 4}, then |H| = 3, and0 +H = {0 + 0, 0 + 2, 0 + 4} = {0, 2, 4} = H = 2 +H = 4 +H,1 +H = {1 + 0, 1 + 2, 1 + 4} = {1, 3, 5} = 3 +H = 5 +H.So G/H = {H, 1 +H} = {{0, 2, 4}, {1, 3, 5}}, and thus |G/H| = 2.Hence indeed |G/H||H| = 2 · 3 = 6 = |G|.

b) If H = 〈3〉 = {0, 3}, then |H| = 2;also 0 +H = H = 3 +H, 1 +H = {1, 4} and 2 +H = 5 +H = {2, 5}.So G/H = {H, 1 +H, 2 +H} = {{0, 3}, {1, 4}, {2, 5}}, and thus |G/H| = 3.In all |G/H||H| = 3 · 2 = 6 = |G| also in this case.

Corollary: If G is a finite group, and |G| is a prime number, then {e} and G are the onlysubgroups of G.

As seen before, if G is a finite group, and |G| is a prime number, then even G = 〈a〉 for everya ∈ G\{e}.

As an application of Lagrange’s theorem we now prove:

Proposition (Classification of groups of order 4): Let G be a group with |G| = 4.

a) If G is cyclic, then G ∼= Z4.

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b) If G is not cyclic, then G ∼= Z2 × Z2.

Proof. Only b) is to be proved, for a) is an instance of the classification of cyclic groups doneearlier. Let G = {e, a, b, c}. By Lagrange, ord(a) =ord(b) =ord(c) = 2, as order can only be 1 or2 and the only element with order 1 is e.

Step 1: We show that ab = c. In fact, ab 6= a and ab 6= b for otherwise b = e or a = e respectively.Also ab 6= e for otherwise b = a−1 and thus {e, a, b} would form a subgroup of G, with order3, which is impossible by Lagrange. So we know that G = {e, a, b, ab}.

Step 2: Showing that ba = c is similar to the argument in Step 1.

Step 3: We conclude that ab = ba.

Step 4: f : Z2 × Z2 → G, (k, `) 7→ akb` is clearly bijective, and f is a homomorphism as follows:For any (ki, `i) ∈ Z2×Z2 where i ∈ {1, 2}, let (k, l) ∈ Z2×Z2 be such that k1+k2 = k mod 2,`1 + `2 = `mod2. Now f((k1, `1) + (k2, `2)) = f(k, l) = akb` = ak1+k2bl1+l2 = ak1bl1ak2bl2 =f(k1, l1)f(k2, l2) since a2 = e = b2 and ab = ba. Hence f is a homomorphism.

Hence G ∼= Z2 × Z2.

10 Permutation Groups

Let n ≥ 1 and write Sn for the symmetric group Sym({1, ..., n}). For πσ ∈ Sn recall that πσ = π◦σ,i.e. πσ(i) = π(σ(i)) for for 1 ≤ i ≤ n. Warning: Sn is not abelian unless n ≤ 2 (see below).

We use small Greek letters π (pi), σ (sigma), τ (tau), and ρ (rho) for elements of Sn, and write(1 2 · · · n− 1 n

σ(1) σ(2) · · · σ(n− 1) σ(n)

)to denote σ ∈ Sn.

Example: n = 3. Claim: S3 is not abelian. Here is a counterexample: if

σ =

(1 2 32 1 3

), τ =

(1 2 31 3 2

),

then σ ◦ τ 6= τ ◦ σ, for e.g. σ ◦ τ(2) = σ(τ(2)) = σ(3) = 3 but τ ◦ σ(2) = τ(σ(2)) = τ(1) = 1.Note that |S3| = 3! = 6, so S3 is the smallest non-abelian group.

Definition The permutations of {1, ..., n} are the elements of Sn. A transposition is a permu-tation τ ∈ Sn for which there are i < j in {1, ..., n} such that

τ(k) =

j if k = ii if k = jk if k 6∈ {i, j}

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If τ is the transposition exchanging i and j as above, then one writes

(i j)

for τ . If τ is a transposition, then τ 2 = τ ◦ τ is the identity map id, the neutral element of Sn;whence τ−1 = τ .

Examples:

1) S1 = Sym({1}) = {id},

2) S2 = Sym({1, 2}) = {id, (1 2)},

3) S3 = Sym({1, 2, 3}) =

{id, (1 2), (1 3), (2 3),

(1 2 32 3 1

),

(1 2 33 1 2

)}

Theorem: |Sn| = n!

Proof. Using Lagrange’s Theorem. We shall use proof by induction.Basis: n = 1. Then |S1| = 1Induction step: Assume result for n. Consider n + 1. To show that |Sn+1| = (n + 1)!, considerH = {σ ∈ Sn+1 : σ(n + 1) = n + 1}. This H is a subgroup of Sn+1, and H ∼= Sn. So |H| = n!by our induction hypothesis. Set G = Sn+1. By Lagrange, |G| = |H||G/H| = n!|G/H|. Since(n+ 1)! = n!(n+ 1), to show that |G| = (n+ 1)! we only have to show that |G/H| = n+ 1. To doso, set τi = (i n+ 1) for any i ∈ {1, ..., n}. We proceed with three claims:

Claim 1) G/H = {H} ∪ {τ1H, ..., τnH}.Proof: For σ ∈ G, set i = σ(n+ 1).If i = n+ 1 then σ ∈ H, i.e. σH = H.If i ≤ n, then τiσ ∈ H (for τiσ(n+1) = τi(i) = n+1), but τ 2

i =id and thus σ = τi(τiσ) ∈ τiH,i.e. σH = τiH.

Claim 2) i 6= j ⇒ τiH 6= τjH where i, j ≤ n.Proof: τjτi(n+ 1) = τj(i) = i 6= n+ 1⇒ τjτi 6∈ H, i.e. τiH 6= τjH.

Claim 3) τiH 6= H for every i ≤ n.Proof: τi 6∈ H, for τi(n+ 1) = i 6= n+ 1.

In all, |G/H| = n+ 1 as required.

Proposition (Classification of groups of order 6): If G is a group of order 6 then

1) If G is cyclic then G ∼= Z6

2) If G is not cyclic then G ∼= S3

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Proof. Using Lagrange’s theorem, the proof is similar to the case of order 4.

Note that Z 6∼= S3, as Z6 is cyclic, but S3 is not even abelian.

Example: The symmetry group D3 of an equilateral triangle is D3 = {I, R, S,A,B,C}, where Idenotes the identity map; R, S the two rotations (R clockwise and S anticlockwise) through 2π

3;

and A,B,C the three reflections through the medians. This D3 cannot be a cyclic group as itdoesn’t have an element of order 6:

ord(I) = 1, ord(R) = ord(S) = 3, ord(A) = ord(B) = ord(C) = 2.

Thus D3∼= S3. In fact, every T ∈ D3 is uniquely determined by its action on the vertices v1, v2, v3

of the triangle, and so by a permutation σ ∈ Sn of their index set {1, 2, 3}:

T (vi) = vj ⇐⇒ σ(i) = σ(j) .

Let again n ≥ 1 and Sn = Sym({1, ..., n}). Then for any σ ∈ Sn and i ∈ {1, ..., n}, if σ(i) 6= i,then σ2(i) 6= σ(i), so σ3(i) 6= σ2(i), etc. But σk(i) = i for some k ≤ n!, e.g. for k = ord(σ).

Definition: The subset Bσ(i) = {σ`(i) : ` ≥ 0} of {1, ..., n} is the orbit of i for σ.

We always have i = id(i) = σ0(i) ∈ Bσ(i), and Bσ(i) = {i} precisely when σ(i) = i.

Lemma: For i, j ∈ {1, ..., n} and σ ∈ Sn we have Bσ(i) ∩Bσ(j) 6= ∅ ⇔ Bσ(i) = Bσ(j).

Proof. Left as an exercise.

Definition: A permutation σ ∈ Sn is a cycle if there is exactly one orbit for σ which has at least2 elements.

Let Bσ(i) be this orbit for σ, so σ(j) = j for every j ∈ {i, ..., n}\Bσ(i) and σ(i) 6= i. If

Bσ(i) = {i, σ(i), σ2(i), ..., σk−1(i)} ,

then k = |Bσ(i)| is the length of the cycle σ, and one writes:

σ = (i σ(i) σ2(i) ... σk−1(i)),

i.e. σ = (i0 i1 i2 ... ik−1) where i0 = i and i`+1 = σ(i`) for every ` < k − 1.

Examples

1) The cycles of length 2 are just the transpositions τ = (i j).

2) In S3 we have two cycles of length 3:

(1 2 3) = (2 3 1) = (3 1 2) =

(1 2 32 3 1

),

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(1 3 2) = (3 2 1) = (2 1 3) =

(1 2 33 1 2

).

Further, the

{cycles of length 3

transpositions

}in S3 correspond to the

{rotations through 2π

3

reflections

}in D3.

Definition: Permutations π, σ ∈ Sn are said to be disjoint if there is no i ∈ {1, ..., n} such thatπ(i) 6= i and σ(i) 6= i.

Proposition:

a) If π, σ ∈ Sn are disjoint, then πσ = σπ,

b) If σ ∈ Sn is a cycle of length k, then ord(σ) = k.

Lemma: Let G be a finite group. If x, y ∈ G such that xy = yx then ord(xy) =lcm(ord(x),ord(y)).

Proof. If xy = yx, then (xy)` = x`y`. The rest is left as an exercise.

Example: In S10, the permutation

(1 7 6 2 3 9)(5 4 8 10)

has order 12: the cycles are disjoint and have order 6 and 4, respectively, for which lcm(6, 4) = 12.

Theorem (Cycle Decomposition of Permutations): For every π ∈ Sn there are disjoint cyclesσ1, ..., σk ∈ Sn with k ≥ 0 such that π = σ1σ2...σk, and this decomposition is unique up torearrangement of σ1, ..., σk.

Proof. Separated into two parts:Existence: Let B1, ..., Bk be the orbits for π which have at least 2 elements. By leaving out repeatedoccurences, we can assume that they are disjoint. For ` ≤ k, define σ` ∈ Sn by

σ`(i) =

{π(i) if i ∈ B`

i otherwise

}.

Then σ` is a cycle, andBσ(i) = B` for every i ∈ B`. Further, σ1, ..., σ` are disjoint and π = σ1σ2...σk.Uniqueness: omitted.

Note that the above existence proof contains an algorithm to compute the cycle decomposition ofa permutation π ∈ Sn: first find the disjoint orbits of π, and then define the cycles by the actionof π on the orbits with ≥ 2 elements.

Lemma: Every cycle can be written as a product of transpositions. In fact

(i0 i1 ... ir−1) = (i0 i1)(i1 i2)...(ir−3 ir−2)(ir−2 ir−1) .

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For example: (3 1 4 2) = (3 1)(1 4)(4 2).

Corollary: For every π ∈ Sn there are transpositions τ1, ..., τq ∈ Sn such that π = τ1...τq.

Examples: We work in S6, a group with 6! = 720 elements.

0) id is the product of q = 0 transpositions, but also

id = (1 2)(1 2) = (3 5)(3 5) = ... = (1 2)(1 2)(1 2)(1 2) = . . .

We notice that in each of these decompositions the number of factors is even.

1) π1 =

(1 2 3 4 5 63 5 2 4 1 6

)has orbits {1, 3, 2, 5}, {4} and {6}, so π1 = (1 3 2 5) is a cycle.

As a product of transpositions, π1 = (1 3)(3 2)(2 5).

2) π2 =

(1 2 3 4 5 63 5 1 4 2 6

)= (1 3)(2 5).

Definition: The inversion number inv(π) of π ∈ Sn is defined by

inv(π) = |{(i, j) ∈ {1, ..., n} : i < j&π(i) > π(j)}| ;

and π is called an

{evenodd

}permutation if inv(π) is an

{evenodd

}number.

Examples:

0) inv(id)= 0, so id is even.

1) Every transposition is odd: if τ = (i j) then inv(τ) = 1 and so τ is odd.

2) In S4, (1 2)(3 4) =

(1 2 3 42 1 4 3

)has inv= 2, and so is even.

3) In S9,

(1 2 3 4 5 6 7 8 96 4 7 2 5 1 8 9 3

)= (1 6)(2 4)(3 7 8 9) is odd.

Definition The sign (or signature) sgn(π) of π ∈ Sn is (−1)inv(π).

Proposition: π is

{evenodd

}if and only if sgn(π) =

{1−1

}.

In particular, sgn(id)=1 and sgn((i j)) = −1.

Lemma: For every π ∈ Sn,

sgn(π) =∏

1≤i<j≤n

i− jπ(i)− π(j)

.

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Proof. Will be inserted later.

Theorem: sgn: Sn → {+1,−1}, π 7→ sgn(π) is a homomorphism.

Proof. Will be inserted later.

Example: If τ is a transposition, and π ∈ Sn, then sgn(τπ) = −sgn(π) =sgn(πτ).

Corollary: Let π ∈ Sn. If π = τ1...τq where τ1, ..., τq are transpositions, then sgn(π) = (−1)q.

In other words, π ∈ Sn is an

{evenodd

}permutation precisely when π is the product of an

{evenodd

}number of transpositions.

Proof. Use induction on q ≥ 0.

Proposition: Every cycle of

{even lengthodd length

}is an

{oddeven

}permutation.

Proof. See problem set 3, question 3a).

From now on, let n ≥ 2.

Definition: The alternating group An is the set {π ∈ Sn : π is even} seen as a subgroup of Sn.

This An actually equals the kernel Ker(sgn)= {π ∈ Sn : sgn(π) = +1} of the homomorphismsgn: Sn → {+1,−1}.

Proposition:

a) |Sn/An| = 2

b) |An| = n!2

Proof. Part (b) follows from part (a) by Lagrange; part (a) will be a by-product of the Homomor-phism Theorem yet to be done.

Examples

1) S2 = {id, (1 2)}, then A2 = {id}.

2) S3 = {id, (1 2), (1 3), (2 3), (1 2 3), (1 2 3)}, thenA3 = {id, (1 2 3), (1 3 2)}.

For more examples see problem set and workshop 3.

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11 Normal Subgroups

Definition: Let G be a group. The conjugacy class of a ∈ G is

[a] = {x−1ax : x ∈ G} .

Note that a ∈ [a] (this is the case x = e), and [e] = {e}. Further, if b ∈ [a] then [b] ⊆ [a].

Definition: A subgroup H of G is normal if for every a ∈ G we have

a ∈ H ⇒ [a] ⊆ H .

We write H CG to say that H is a normal subgroup of G.

Note that {e}CG and GCG: the trivial subgroups are normal.

Proposition:

a) If G is abelian, then every subgroup of G is normal.

b) If f : G → G′

is a homomorphism, where G and G′

are arbitrary groups, then Ker(f) ={x ∈ G : f(x) = e

′} is a normal subgroup of G.

Proof. a) If G is abelian then [a] = {a}, for if b ∈ [a], say b = x−1ax with x ∈ G, thenb = x−1xa = a, hence if H is a subgroup of G, and a ∈ H, then [a] = {a} ⊆ H.

b) Let a ∈ Ker(f), i.e. f(a) = e′. If b ∈ [a], say b = x−1ax with x ∈ G, then

f(b) = f(x−1ax) = f(x−1)f(a)f(x) = f(x)−1e′f(x) = f(x)−1f(x) = e

′

so b ∈ Ker(f). This proves [a] ⊆ Ker(f).

Examples:

1) Z C R where R stands for the group (R,+, 0), which is abelian.

2) SLn(K) CGLn(K) for K ∈ {Q,R,C}, as SLn(K) = Ker(det) where

det : GLn(K)→ K∗, A 7→ det(A)

is a homomorphism.

3) AnCSn where n ≥ 2, because An = Ker(sgn) and sgn: Sn → {+1,−1} is a homormorphism.

23

Definition: Let G be any group and a ∈ G. The centraliser CG(a) of a ∈ G is

{x ∈ G : ax = xa} .

Note that ax = xa precisely when x−1ax = a.

Lemma: CG(a) is a subgroup of G with |G/CG(a)| = |[a]|.

Proof. Omitted

Example: S3 = {id, (1 2), (1 3), (2 3), (1 2 3)(1 3 2)} has three conjugacy classes:

{id}, {(1 2), (1 3), (2 3)}, {(1 2 3), (1 3 2)}.

To see this, it may help to know the number of elements of each conjugacy class—or, equivalently,the orders of the centralisers. First, CG(id) = S3, so |[id]| = |S3/S3| = 1, and [id] = {id}. Next,

CG((1 2)) = {id, (1 2)}

(check this!); whence, by Lagrange,

|[(1 2)]| = |S3/CG((1 2))| = |S3||CG((1 2))|

=3!

2= 3

and likewise for (1 3) or (2 3) in place of (1 2). These three transpositions belong to the sameconjugacy class (check this!): that is,

[(1 2)] = [(1 3)] = [(2 3)] = {(1 2), (1 3), (1 2)}.

Last,CG((1 2 3)) = {id, (1 2 3), (1 3 2)} ;

whence, by Lagrange,

|[(1 2 3)]| = |S3/CG((1 2 3))| = |S3||CG((1 2 3))|

=3!

3= 2

and likewise for (1 3 2) in place of (1 2 3). These two cycles of length 3 belong to the sameconjugacy class (check this!): that is,

[(1 2 3)] = [(1 3 2)] = {(1 2 3), (1 3 2)}.

For the (normal) subgroups of S3 see the worked example that is available on the VLE.

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